Show that if the diagonals of a quadrilateral bisect eachother at right angels then it is a rhombus
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We have a quadrilateral ABCD such that the diagonals AC and BD bisect each other at right angles at O.
∴ In ΔAOB and ΔAOD, we have
AO = AO
[Common]
OB = OD
[Given that O in the mid-point of BD]
∠AOB = ∠AOD
[Each = 90°]
ΔAOB ≌ ΔAOD
[SAS criteria]
Their corresponding parts are equal.
AB = AD
...(1)
Similarly,
AB = BC
...(2)
BC = CD
...(3)
CD = AD
...(4)
∴ From (1), (2), (3) and (4), we have AB = BC CD = DA
Thus, the quadrilateral ABCD is a rhombus.
∴ In ΔAOB and ΔAOD, we have
AO = AO
[Common]
OB = OD
[Given that O in the mid-point of BD]
∠AOB = ∠AOD
[Each = 90°]
ΔAOB ≌ ΔAOD
[SAS criteria]
Their corresponding parts are equal.
AB = AD
...(1)
Similarly,
AB = BC
...(2)
BC = CD
...(3)
CD = AD
...(4)
∴ From (1), (2), (3) and (4), we have AB = BC CD = DA
Thus, the quadrilateral ABCD is a rhombus.
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